TI-84 Plus Graphing Calculator is required for this course. · Multiplying Polynomials and Solving...

Summer Review for Students Entering Algebra 2

1. Solving Linear Equations

2. Solving Linear Systems of Equations

3. Multiplying Polynomials and Solving Quadratics

4. Writing the Equation of a Line

5. Laws of Exponents and Scientific Notation

6. Solving with Trig Ratios and Pythagorean Theorem

TI-84 Plus Graphing Calculator is required for this course.


TOPIC 1: Solving Linear Equations

Equations in one variable with fractional coefficients may be solved in a variety of ways. Commonly, the first

step is to multiply all the terms by a common denominator to remove the fractions. Then solve in the usual way.

Combine like terms. Isolate the variable on one side and the constants on the other. Finally, divide to find the

value of the variable.

Example 1: Solve x + x − 3 = x + 4

Example 2: Solve for y: + − 3 = y

Solve the following equations.

1. x − 7 = x + 3 2. x − 7 = x – 5 3. y = y + 7

4. x + 7 = − x + 3 5. y + 7 + y = y – 1 6. + + x = 2

Answers: 1. x = 30 2. x = 8 3. y = 84 4. x = − 5. no solution 6. x =


TOPIC 2: Solving Linear Systems of Equations

When two equations are both in y = mx + b form it is convenient to use the Equal Values Method to solve

for the point of intersection. Set the two equations equal to each other to create an equation in one variable

and solve for x. Then use the x-value in either equation to solve for y.

If one of the equations has a variable by itself on one side of the equation, then that expression can replace

the variable in the second equation. This again creates an equation with only one variable. This is called

the Substitution Method. See Example 1 below.

If both equations are in standard form (that is ax + by = c), then adding or subtracting the equations may

eliminate one of the variables. Sometimes it is necessary to multiply before adding or subtracting so that

the coefficients are the same or opposite. This is called the Elimination Method. See Example 2 below.

Sometimes the equations are not convenient for substitution or elimination. In that case one of both of the

equations will need to be rearranged into a form suitable for the previously mentioned methods.

Example 1: Solve the following system. 4x + y = 8

x = 5 − y

Solution: Since x is alone in the second equation, substitute 5

− y in the first equation, then solve as usual.

Then substitute y = 4 into either original equation to

find x. Using the second equation x = 1 so the solution is (1, 4)

x = 5 − 4

x = 1

Example 2: Solve the following system. −2x + y = −7

3x − 4y = 8

Solution: If we add or subtract the two equations no variable is eliminated. Notice, however,

that if everything in the top equation is multiplied by 4, then when the two equations are

added together, the y-terms are eliminated.


Substituting x = 4 into the first equation −2(4) + y = −7 ⇒ −8 +y = −7⇒ y = 1.

The solution is (4,1).

Solve the following systems of equations.

1. y = 7x − 5

2x + y = 13

2. x + y = −4

−x + 2y = 13

3. y = −3x

4x + y = 2

4. y − x = 4

2y + x = 8

5. 2x − y = 4

x + y = 1

6. −4x + 6y = −20

2x − 3y = 10

Answers:

1. (2, 9) 2. (−7, 3) 3. (2, −6) 4. (0, 4) 5. (2, 0) 6. infinite

solutions

TOPIC 3: Multiplying Polynomials and Solving Quadratics

Polynomials can be multiplied (changed from the area written as a product to the area written as a

sum) by using the Distributive Property or generic rectangles.

Example 1: Multiply −5x(−2x + y).

Solution: Using the Distributive Property −5x(−2x + y) = −5x • −2x + −5x • y = 10x2 − 5xy

area as a product area as a sum

Example 2: Multiply (x − 3)(2x + 1).


Solution: Although the Distributive Property may be used, for this problem and other more

complicated ones, it is beneficial to use a generic rectangle to find all the parts.

⇒ (x − 3)(2x + 1) = 2x2 − 5x − 3

area as a product area as a sum

Solving quadratics first required factoring the polynomials to change the sum into a product. It is

the reverse of multiplying polynomials and using a generic rectangle is helpful. Once the

expression is factored, then the factors can be found using the Zero Product Property.

Example 3: Solve x2 + 7x + 12 = 0

Solution: First, factor the polynomial. Sketch a

generic rectangle with 4 sections.

Write the x2 and the 12 along one diagonal.

Find two terms whose product is 12 • x2 =

12x2 and whose sum is 7x. That is, 3x and 4x.

This is the same as a Diamond Problem.

Write these terms as the other diagonal.

Find the base and height of the rectangle by

using the partial areas.

Write the factored equation. x2 + 7x +12 = (x +

3)(x + 4) = 0

Then, using the Zero Product Property, we know

that either (x + 3) or (x + 4) is equal to zero

(since their product is zero). This means

that x = −3 or x = −4.


Here are some more to try. Multiply the expression in problems 1 through 4 and solve the equation

in problems 5 and 6.

1. 2x(x − 1) 2. (3x + 2)(2x + 7) 3. (2x − 1)(3x + 1)

4. (x + y)(x + 2) 5. x2 + 5x + 6 = 0 6. 2x2 + 5x + 3 = 0

Answers:

1. 2x2 − 2x 2. 6x2 + 25x + 14 3. 6x2 − x − 1

4. x2 + xy + 2x + 2y 5. x = −2, −3 6. x = − , −1

TOPIC 4: Writing the Equation of a Line

Except for a vertical line, any line may be written in the form y = mx + b where “b” represents the y-intercept of

the line and “m” represents the slope. (Vertical lines are always of the form x = k .) The slope is a ratio

indicating the steepness and direction of the line. The slope is calculated by .

Example 1: Write the equation of the line with slope − and passing through the point (6, 3).

Solution: Write the general equation of a line.

Substitute the values we know for m, x, and y.

Solve for b.

Write the complete equation.

y = mx + b

3 = − (6) +b

3 = −3 + b

6 = b

y = − x + 6


Example 2: Write the equation of the line passing through the points (8, 3) and (4, 6).

Solution: Draw a generic slope triangle

Calculate the slope using the given two points.

Write the general equation of a line.

Substitute m and one of the points for x and y,

in this case (8,3).

Solve for b.

Write the complete equation.

y = mx + b

3 = − (8) + b

3 = −6 + b

9 = b

y = − x + 9

Use the given information to find an equation of the line.

1. slope = 5, through (3, 13)

2. slope = − , through (3, −1) 3. slope = , through (6, 9)

4. through (1, 3), (−5, −15) 5. through (2, −1), (3, −3) 6. through (1, −4), (−2, 5)

7. 8.

9.

.

Summer Review for Students Entering Algebra 2 Answers:

1. y = 5x − 2

2. y = − x + 4 3. y = x + 7

4. y = 3x 5. y = −2x + 3 6. y = −3x – 1

7. y = 2x + 1 8. y = x + 3 9. y ≈ x − 4

TOPIC 5: Laws of Exponents and Scientific Notation

The laws of exponents summarize several rules for simplifying expressions that have exponents. The rules are

true for any base if x ≠ 0 .

xa · x

b = x

(a+b)

x0 = 1

(xa)b = x

ab

= x(a−b)

Scientific notation is a way of writing a number as a product of two factors separated by a multiplication sign.

The first factor must be less than 10 and greater than or equal to 1. The second factor has a base of 10 and an

integer exponent.

Example 1: Simplify 42 · 4−4

Solution: In a multiplication problem, if the bases are the same, add the exponents and keep

the base. If the answer ends with a negative exponent, take the reciprocal and change the

exponent to positive.

42 · 4−4 = 4−2 = =


Example 2: Simplify

Solution: Separate the fraction into two fractions with bases x and y. With an exponent on an

exponent, multiply the exponents. Next, to divide expressions with exponents and the same

base, subtract the exponents.

Example 3: Multiply and give the answer in scientific notation. (8 × 104) · (4.5 × 10−2)

Solution: Separate the number parts and the exponent parts. Multiply the number parts

normally and the exponent part by adding the exponents. If this answer is not in scientific

notation, change it appropriately.

(8 × 104) · (4.5 × 10−2) = 8 × 4.5) · (104 × 10−2) = 36 × 102 = (3.6 × 101) × 102 = 3.6 × 103

Simplify each expression. For problems 7 and 8 write the final answer using scientific notation.

1. 33 · 3

3 · 3

6 2. 5

4 · 5

−1 3. x

2 · (x

4)−2

4.

5. (72 · 7

3)4

6.

7. (4.25 × 103) · (2 × 10

5) 8. (1.2 × 10

4) · (7.1 × 10

-2)

Answers:

1. 312

2. 53

3. x−6

= 4. y

4 5. 7

20 6. y

3 7. 8.5 × 10

8 8. 8.52 × 10

2

Summer Review for Students Entering Algebra 2 TOPIC 6: Solving With Trigonometric Ratios and the Pythagorean Theorem

Three trigonometric ratios and the Pythagorean Theorem can used to solve for the missing side

lengths and angle measurements in any right triangle.

In the triangle below, when the sides are described relative to the angle θ, the opposite leg is y and

the adjacent leg is x. The hypotenuse is h regardless of which acute angle is used.

Also for the triangle above, from the Pythagorean Theorem: h2 = x2 + y2.

Example 1: A rectangle has a diagonal of 16 cm and one side of 11 cm. What is the length of the

other side?

Solution: The diagonal represents the hypotenuse of a right triangle and the one given side

represents one of the legs. Using the Pythagorean Theorem:

h2 = x2 + y2 ⇒162 = 112 + y2 ⇒ 256 = 121+ y2

135 = y2 ⇒ y = ≈ 11.62 cm

Example 2: Solve for x in the triangle at right.

Solution: Based on the 40º angle, 11 is the adjacent side and x is the hypotenuse. Use the cosine


ratio to solve.

Example 3: A ten-foot ladder is leaning against the side of a house. If the top of the ladder touches

the house nine feet above the ground, what is the angle made by the ladder and the ground?

Solution: Make a diagram of the situation similar to the one at right. The ladder (10) is the

hypotenuse and the house (9) is the opposite leg. Using the sine ratio, sinθ = To find θ, “undo”

the sine function with the inverse sine function (sin−1 x) as follows:

Here are some more to try. Use the right triangle trigonometric ratios or the Pythagorean

Theorem to solve for the variable(s).

1.

2.

3.

4.


5.

6.

7.

8.

9.

Draw a diagram and solve the following.

10. What is the distance between (−6, −6) and (−3, 2)?

Answers:

1. 5 units 2. ≈ 50.7 units 3. ≈ 10.4 units

4. 5 units 5. 6. 3 units

7. 8. ≈ 48.2° 9. 60°

10.

TI-84 Plus Graphing Calculator is required for this course. · Multiplying Polynomials and Solving...

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